Homework Help:
Finding the Inverse of an Epimorphism

1. The problem statement, all variables and given/known data
Let ##f : G \rightarrow H## be an epimorphism from a group ##G## to ##H## and let ##h \in H##, then ##f^{-1} (h) = g ~ker(f)##.

2. Relevant equations

3. The attempt at a solution
So, if I understand the problem correctly, we are trying to find a epimorphism which has a rule such that its inverse has the rule ##f^{-1} (h) = g~ker(f)##, which essentially says that ##f^{-1}## maps an element in ##H## to a left-coset of ##ker(f)##. In this light, I am rather confused why they say that ##f## maps between the groups ##G## and ##H##, rather than ##G/g~ker(f)## and ##H##.

I am having a rather difficult time finding such a rule for the epimorphsim ##f##, so that ##f^{-1} (h) = g~ker(f)##.

Let ##f : G \rightarrow H## be an epimorphism from the group ##G## to ##H## and let ##h \in H##, then ##f^{-1}(h) = g ~ker(f)## for some ##g \in G##.

By the way, I am slightly confused as to what ##f^{-1} (h) = g ~ker(f)## even means. What is the rule of assignment? Obviously, ##h## gets mapped to ##g~ker(f)##, but what would, say, ##h_1## get mapped to? ##g_1 ~ker(f)##? It does not seem clear.

Here is another thing with which I am concerned. Isn't ##f^{-1}## mapping an element in ##H## to an element in ##G/g~ker(f)##? And is it not true that ##f## and ##f^{-1}## have to be acting over the same sets? If we are letting ##f : G \rightarrow H##, it would not seem that they are acting over the same sets.

but what would, say, ##h_1## get mapped to? ##g_1 ~ker(f)##? It does not seem clear.

You should form the habit of getting your variables like [itex] h_1, g_1 [/itex] within a proper "scope". If you are familiar with computer programming, a variable like "x" in a computer program might have one meaning in one function and another meaning in another function. The syntax checking of the compiler forces you to put a variable like "x" within some "scope". If you don't, it tells you the variable is "undefined". The scope of variables in written mathematics is done less formally, but the effect must still be there to make the writing clear. The scope of a variable like [itex] h [/itex] is usually defined by a quantifier, such as "for each" or "there exists". In your problem the phrase "for some [itex] h [/itex] " is used instead of "there exists an [itex] h [/itex]". English has a great variety of phrases the have the same meaning when interpreted as pure logic.

How are you using the variable [itex] h_1 [/itex]? What defines its scope? Do you want to say "Let [itex] h_1 [/itex] be an arbitrary element of [itex] H [/itex]"? When we want to make a statement of the form "For each [itex] h_1 \in H. [/itex] ...." we can say that in English by the phrase "Let [itex] h_1 [/itex] be an arbitrary element of [itex] H [/itex]".

Defining the scope of variables in your writing, will help you think clearly. Using an extensive vocabulary is good style in writing fiction, but in writing mathematics, you get a clearer result by using a limited vocabulary. Take for example, your use of the word "acting". There are technical definitions for the relation "acts on" in various mathematical contexts. You haven't explained what you mean by "acting".

To make the "for each" and "there exists" logical quantifiers clear we can say "Let [itex] h [/itex] be an arbitrary element of [itex] H [/itex]. There exists at least one element [itex] g [/itex] such that [itex] f(g) = h [/itex] since [itex] f [/itex] is an epimorphism." Proceed to prove 1) and 2).

Okay, so in post #9 I demonstrated that ##g ~ker(f) \subset f^{-1}(h)## by showing that some arbitrary element in ##g~ker(f)## gets mapped to the element ##h##, implying that the element is also in ##f^{-1}(h)##.

Now, I am trying to prove the statement 1) you mentioned in post #10. Here is my work:

Let ##x \in f^{-1}(h)## be arbitrary. Then by definition of the preimage, this implies that ##f(x) = h##. Let us form the left-coset of ##ker(f)## from the element ##x##: ##x~ker(f)##. We know that ##x \in x ~ker(f)##. Therefore, the arbitrary element ##x## in ##f^{-1}(h)## was just shown to be in ##x~ker(f)## also. Therefore, ##f^{-1} \subset x~ker(f)##.

Okay, so in post #9 I demonstrated that ##g ~ker(f) \subset f^{-1}(h)## by showing that some arbitrary element in ##g~ker(f)## gets mapped to the element ##h##, implying that the element is also in ##f^{-1}(h)##.

You might have shown that, depending on how you intend to incorporate the symbol [itex] g [/itex] in that work.

Now, I am trying to prove the statement 1) you mentioned in post #10. Here is my work:

Let ##x \in f^{-1}(h)## be arbitrary. Then by definition of the preimage, this implies that ##f(x) = h##. Let us form the left-coset of ##ker(f)## from the element ##x##: ##x~ker(f)##.

The left coset [itex] x\ ker(f) [/itex] isn't relevant. You need to show [itex] x [/itex] is in the left coset [itex] g\ ker(f) [/itex].
The symbol [itex] g [/itex] is used earlier in the proof to denote a specific element.. You should give an argument that is still within the "scope" of that use of [itex] g [/itex].

It's worth remembering that, in a group, first degree equations always have a unique solution. For example, to solve for [itex] y [/itex] in the equation [itex] ay = b [/itex] you can multiply both sides on the left by [itex] a^{-1} [/itex] obtaining [itex] y = a^{-1} b [/itex].

As a consequence of this, if you are discussing a element [itex] h [/itex], you can always force another element [itex] g [/itex] into the discussion by factoring [itex] h [/itex] into a product invovling [itex] g [/itex]. To find the factorization with [itex] g [/itex] on the left, you solve equation [itex] h = gy [/itex] for [itex] y [/itex] obtaining [itex] y = g^{-1}h [/itex] and [itex] h = g( g^{-1} h) [/itex].

We have assumed [itex]x \in f^{-1}(h) [/itex]. This implies [itex] f(x) = h [/itex]
Using the above factorization, [itex] f(x) = h = g (g^{-1}x) [/itex] This has the form [itex] g [/itex] times something. To show [itex] x \in g\ ker(f) [/itex] you need to show the factor [itex] g^{-1} x [/itex] is in [itex] ker(f) [/itex]. To do that you need to show [itex]f(g^{-1} x) [/itex] is the identity. Show that by showing it is equal to [itex]h^{-1} h [/itex].